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03 Sep 2019, 11:53
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Difficulty:
95%
(hard)
Question Stats:
41% (02:13) correct
59%
(02:23)
wrong
based on 82
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How many numbers with two or more digits can be formed using the digits 1, 2, 3, 4, 5, 6 and 7 so that in every such number, each digit is used at most once and the digits appear in the ascending order?
A. 6
B. 56
C. 57
D. 120
E. 121
A. 6
B. 56
C. 57
D. 120
E. 121
05 Sep 2019, 08:32
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Dillesh4096
Irrespective of the number of digits, there is ONLY one way in which the digits can be arranged in ascending order
This means COMBINATION is what we are looking at..
2 digits would therefore mean 7C2 and similarly others..
So our answer is \(7C2+7C3+7C4+7C5+7C6+7C7 = \frac{7*6}{2}+\frac{7*6*5}{3*2}+\frac{7*6*5}{3*2}+\frac{7*6}{2}+7+1=21+35+35+21+7+1=120\)
A quick way is \(7C0+7C1+7C2+7C3+7C4+7C5+7C6+7C7 -(7C0+7C1)=2^7-(1+7)=128-8=120\)
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05 Sep 2019, 18:21
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As the digits are in ascending order, there is only 1 arrangement of digits possible.
Total numbers with two or more digits can be formed= [7C2+7C3+7C4+7C5+7C6+7C7]*1= {\(2^7-7-1\)}= 120
Total numbers with two or more digits can be formed= [7C2+7C3+7C4+7C5+7C6+7C7]*1= {\(2^7-7-1\)}= 120
Dillesh4096
